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       	\author{J.~Gladines, G.~Mertes}
	\title{Lock-in Amplifier}
	
       	\begin{document}
      		\maketitle
		
		\section{Introduction}
		To understand the workings of a lock-in amplifier, we must first understand some of the mathematics behind it. In this chapter we will give an introduction of the mathematical principals on which a lock-in 
		amplifier is based.
		
		\subsection{the Fourier transform}
		When working with signals, one must understand that signals can be defined in both the time and frequency domain. For example let's suppose the signal of interest is a sinusoidal function of time. The most common way
		to write this sine function is:
		
		\begin{equation}\label{eq:basic_sine}
				f(t) = A \cdot \sin (\omega \cdot t + \phi)
		\end{equation}

		
		To obtain the definition of the signal as a function of frequency, we need to apply the Fourier transform to \eqref{eq:basic_sine}. This can be accomplished by calculating the Fourier integral\eqref{eq:fourier_transf}.
		The Fourier transform is a very important operation in the analysis of signals and linear, time-invariant systems. By looking at the properties of the Fourier transform, we see why it's so useful\cite{signsyst}. 
		The first property we discuss is it's linearity \cite{fourierbook}. When we scale a function of time, it's transform pair will be scaled in a linear way \eqref{eq:fourier_linearity_1}, or when we add two functions the 
		Fourier transform of the result is equal to the sum of the individual transforms of the functions \eqref{eq:fourier_linearity_2}.
		
		\begin{equation}\label{eq:fourier_transf}
				\mathscr{F}\{f(t)\} = F(f) = \int_{-\infty}^{+\infty} f(t) \cdot e^{-j\omega t }\cdot dt
		\end{equation}
				
		\begin{subequations}\label{eq:fourier_linearity}
			\begin{align}
				\mathscr{F}\{a \cdot g(t)\}& = a\cdot G(f)\label{eq:fourier_linearity_1}\\
				\mathscr{F}\{g(t) + h(t)\}& =  G(f) + H(f)\label{eq:fourier_linearity_2}
			\end{align}
		\end{subequations}
		
		\begin{equation}\label{eq:fourier_tshift}
				\mathscr{F}\{f(t-t_0)\} = e^{j\omega t_0}F(f)		
		\end{equation}
		
		\begin{equation}\label{eq:fourier_fshift}
				\mathscr{F}\{e^{j\omega t_0}f(t)\} = F(f - f_0)		
		\end{equation}
		
		A second important property is time shift \cite{fourierbook}. When a function is shifted in time, a phase shift is induced, proportional to the frequency and the amount of time shift \eqref{eq:fourier_tshift}. 
		Besides the time shift a frequency shift \cite{fourierbook} can also be defined. When the frequency of the transformed function is shifted, the function of time is multiplies with a unit phasor proportional to the frequency and 
		the amount of displacement in frequency \eqref{eq:fourier_fshift}. These are the most important properties of the Fourier transform. To follow up with the example given, it's not easy to calculate the Fourier integral. 
		But if we use Eulers representation of a sine function calculating the integral becomes much easier. In order to simplify the equation some more we will choose $A = 1$ and $\phi = 0$.
		
		
		\begin{subequations}\label{eq:fourier_transf_sine}
			\begin{align}
				\mathscr{F}\{\sin (2 \pi f_0 \cdot t)\}& = \int_{-\infty}^{+\infty} \frac{e^{j 2 \pi f_0 t } - e^{-j 2 \pi f_0 t }}{2j} \cdot e^{-j 2 \pi f t }\cdot dt\\
				& = \frac{1}{2j}\left[ \int_{-\infty}^{+\infty} e^{-j 2 \pi (f + f_0) t } dt - \int_{-\infty}^{+\infty}e^{-j 2 \pi (f - f_0) t }dt \right]\\
				& =  \frac{1}{2j}\left[ \delta(f+f_0) - \delta(f-f_0) \right]
			\end{align}
		\end{subequations}
		
		As you can see the transform pair of a sine function is a combination of 2 dirac delta functions, shifted to the left and the right of zero. The result is visualised in figure~\ref{fig-fourier-sine-transf}
		
		\begin{figure}[htb]
					\center
					\includegraphics[width=65mm]{images/hoofdstuk-1/fourier_1.png}
					\caption{Fourier transform pair of a sine function}
					\label{fig-fourier-sine-transf}
		\end{figure}
		
		Another way to define the Fourier transform is "A family of mathematical techniques based on decomposing signals into
		 sinusoids. In the complex version, signals are decomposed into complex exponentials"\cite{dspguide}. What this means is that almost every signal can be made by adding and subtracting sinusoids. If for example we
		 add the three sinus functions in figure~\ref{fig-fourier-series} we get a signal that approaches a square wave. Due to the linearity of the Fourier transform, the transform pair of the signal is equal to the sum of the 
		 transform pairs of the individual components. Therefore we can easily find the Fourier transform of the signal since we know the transform pair of a sine function.
		
		\begin{figure}[H]
					\center
					\includegraphics[width=155mm]{images/hoofdstuk-1/fourier_2.png}
					\caption{First three components to form a square wave}
					\label{fig-fourier-series}
		\end{figure}
		
		
		
		\subsection{sinusoid multiplication}
		The Fourier series tell us how to decompose signals into their components. Since these components are sinusoids we must understand some mathematics involving sinusoids. More importantly we need to understand 
		what the result is of multiplication of two sinusoids. The result of an equation like $(A_1 \cdot \sin (\omega_1 \cdot t)) \cdot (A_2 \cdot \sin (\omega_2 \cdot t))$ can be found in the collection of trigonometric 
		formulas called Simpsons formulas. The answer to the equation is $\frac{A_1 \cdot A_2 \cdot [cos((\omega_1-\omega_2)\cdot t)-cos((\omega_1 + \omega_2)\cdot t)]}{2}$
		\subsection{convolution}
		
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